问题
sample input= 10 -20 -3916237 -357920 -3620601 7374819 -7330761 30 6246457 -6461594 266854
sample output = -20 30
thanks in advance i am beginner in programing
class Main {
public static void main(String[] args) {
int _a_size = Integer.parseInt(args[0]);
Assert.assert2(args.length,_a_size+1);
int[] _a = new int[_a_size];
for(int _a_i = 0;_a_i<_a_size;_a_i++) {
_a[_a_i] = Integer.parseInt(args[_a_i+1]);
}
Operation.minDiffrence (_a);
}
}
回答1:
You can do
public static List<int[]> minDifference(int[] values) {
Arrays.sort(values); // O(N log N)
List<int[]> results = new ArrayList<>();
long minDiff = Long.MAX_VALUE;
for(int i = 0; i < values.length-1;i++) {
long diff = Math.abs((long) values[i+1] - values[i]);
if (diff < minDiff) {
results.clear();
minDiff = diff;
}
if (diff == minDiff)
results.add(Array.copyOfRange(values, i, 2));
}
return results;
}
What this does is maintain all the results which are the shortest so far, and if a short result is found, it discards the old ones. You get back a list of pairs of equally separated values.
Note: the calculations are performed as long
to avoid overflows.
回答2:
I know this is an old question, but here is a working Java 8 solution:
Map<Integer, List<List<Integer>>> map = list.stream().flatMap(i -> list.stream()
.map(i2 -> (i.equals(i2) || i > i2) ? null : Arrays.asList(i, i2)))
.filter(ints -> ints != null)
.collect(Collectors.groupingBy(o -> Math.abs(o.get(1) - o.get(0))));
It looks a bit ugly in SO's code display, but in my IDE it fits nicely on 3 lines. It returns a map where the keys are the absolute difference and the values are a list of the corresponding pairs of numbers (because there might be more than one).
You might not need it anymore, but it's always fun playing with streams.
来源:https://stackoverflow.com/questions/28189282/integers-which-have-smallest-absolute-difference-there-can-be-one-such-pair-or-m